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The Schroedinger equation for the hydrogen atom is

$$\Big(-\frac{\hbar^2}{2m} \Delta - \frac{e^2}{r} \large)\Psi(\mathbf{r}) = E \; \Psi(\mathbf{r}).$$

I have found that the momentum representation of the above equation reads

$$\frac{\mathbf{p}^2}{2m}\Psi(\mathbf{p}) -\frac{e^2}{2\pi^2 h} \int \frac{\Psi(\mathbf{p'})\mathrm{d}^3\mathbf{p'}}{|\mathbf{p}-\mathbf{p'}|^2} = E \; \Psi(\mathbf{p}).$$ How to derive it? The Fourier transforms of the Laplace operator part and the right hand side are clear. But how about the potential part?

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The problem here is the mistaken impression that the momentum representation is connected to the position representation by a Fourier Transform. Contrary to what is said in textbooks, this is only true for Cartesian Coordinates. For curvilinear coordinates (e.g. r, theta, phi) a different transformation is needed. Check out the article: The Hydrogen Atom in the Momentum Representation, John R. Lombardi, Phys. Rev. A, 22, 797 (1980), in which the correct transform leads to a momentum space where all the momentum variables are chosen to be properly conjugate to the three position space coordinates.

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  • $\begingroup$ We do have MathJax here to format equations. You can see the notation page in help center for details, if you aren't familiar with it (though it's similar to LaTeX). $\endgroup$ – Kyle Kanos Jun 6 at 20:29
  • $\begingroup$ Hi, thank you for the wonderful reference I had not known about. $\endgroup$ – DanielC Jun 6 at 21:22
  • $\begingroup$ I will look at the suggested reference, but I think that these argumemts are not really sound as the equations in OP are not linked with a particular (curvilinear) coordinates system. IMO @Blazej's answer is totality correct. $\endgroup$ – Jhor Jun 11 at 7:46
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You take the Fourier transform of Schrödinger equation in position space. Transformation of the term with potential can be calculated using convolution theorem (Fourier transform of a product is convolution of Fourier transforms - details are easy to find online and the proof is rather straightforward). Fourier transform of $V$ itself can be easily calculated using three facts (I write up to constant factors which can be easily checked):

  1. Fourier transform of Laplace operator acting on function is Fourier transform of that function times $-k^2$ (here $k$ is wavector).

  2. Laplace operator on $V$ is delta.

  3. Fourier transform of delta is a constant.

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  • $\begingroup$ Why do I need Laplace operator on V if Laplace is acting only on the wave function? By using the convolution theorem it seems that what I need is the Fourier transform of 1/r. Then I would use the fact that the Fourier transform of a product is the convolution of the respective Fourier transforms. That should solve the problem. $\endgroup$ – wondering Jan 19 '17 at 12:11
  • $\begingroup$ Yes. I was hinting to the fact that knowing that Laplace operator acting on $1/r$ gives delta you can immediately deduce what is its Fourier transform. $\endgroup$ – Blazej Jan 19 '17 at 13:46

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